What are the solutions to the equation 0=2x^2-4x+2 ?

The solutions to the equation 0=2x^2-4x+2 are x=1

Find the zeros of 0=2x^2-4x+2

The zeros of 0=2x^2-4x+2 are x=1

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0 = 2 x^{2} - 4 x + 2

x = 1

Solution steps

0 = 2 x^{2} - 4 x + 2

Switch sides

2 x^{2} - 4 x + 2 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 2 \cdot 2}{2 \cdot 2}

(- 4)^{2} - 4 \cdot 2 \cdot 2 = 0

x_{1, 2} = \frac{- ( - 4 ) \pm 0}{2 \cdot 2}

x = \frac{- ( - 4 )}{2 \cdot 2}

\frac{- ( - 4 )}{2 \cdot 2} = 1

x = 1

The solution to the quadratic equation is:

x = 1

4 = 2 x^{2} - x + 3

2 t^{2} - 12 t + 16 = 0

(3 k + 2)^{2} = 100

y, x = 4 - (y - 2)^{2}

x^{2} = 1 5^{2} - 1 2^{2}

What are the solutions to the equation 0=2x^2-4x+2 ?
The solutions to the equation 0=2x^2-4x+2 are x=1
Find the zeros of 0=2x^2-4x+2
The zeros of 0=2x^2-4x+2 are x=1